Question

solve the following limit​

Answer 1

Answer:

final answer is 6√5

for step cheak attachment

Answer 2

Answer:

6√5

Step-by-step explanation:

By directly substituting the limits x = 3, we get (0/0) which is an indeterminate quantity.

Let's solve it by using the method of rationalisation.

Rationalising factor for the denominator of given function is $\sqrt{3x-4} + \sqrt{x+2}$, so multiply both numerator and denominator by this factor.

${ \implies \lim \limits_{x \to 3} \dfrac{ {x}^{2} - 9}{ \sqrt{3x - 4} - \sqrt{x + 2} } \times \dfrac{ \sqrt{3x - 4} + \sqrt{x + 2} }{ \sqrt{3x - 4 + \sqrt{x + 2} } } }$

${ \implies \lim \limits_{x \to 3} \dfrac{(x + 3)(x - 3)(\sqrt{3x - 4} + \sqrt{x + 2})}{ (\sqrt{3x - 4})^{2} - ( \sqrt{x + 2} )^{2} } }$

${ \implies \lim \limits_{x \to 3} \dfrac{(x + 3)(x - 3)( \sqrt{ 3x - 4} + \sqrt{x + 2})}{3x - 4 - x - 2 } }$

${ \implies \lim \limits_{x \to 3} \dfrac{(x + 3)(x - 3)( \sqrt{ 3x - 4} + \sqrt{x + 2})}{2x - 6 } }$

${ \implies \lim \limits_{x \to 3} \dfrac{(x + 3) \cancel{(x - 3)}( \sqrt{ 3x - 4} + \sqrt{x + 2})}{2 \cancel{(x - 3)}} }$

${ \implies \lim \limits_{x \to 3} \dfrac{(x + 3)( \sqrt{ 3x - 4} + \sqrt{x + 2})}{2} }$

${ \implies \dfrac{(3+ 3)( \sqrt{ 3(3) - 4} + \sqrt{3 + 2})}{2} }$

${ \implies \dfrac{(6)( \sqrt{9 - 4} + \sqrt{5})}{2} }$

${ \implies \dfrac{6( \sqrt{5} + \sqrt{5})}{2} }$

${ \implies \dfrac{6( 2\sqrt{5})}{2} }$

$\implies6 \sqrt{5}$